## Biochem Help?

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Chemhalp
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### Biochem Help?

What is the concentration in µg ml-1 of a solution of glycine of 23 g litre-1? (2 marks)
(Mol. wt of glycine = 75.07)

Thanks ?

Can I get some pointers on where to start

5. The following colony counts on nutrient agar plates were obtained from dilutions of an overnight culture of the bacterium Pseudomonas aeruginosa:

10-5 = >1000
10-6 = 622
10-7 = 74
10-8 = 14

Choose an appropriate colony count to determine how many colony-forming units (cfu) were present in 1 ml of the original culture (assuming that 100 l of the diluted culture was spread onto each agar plate)? (4 marks)

6. 50 mg of the amino acid isoleucine (molecular weight of 131.2) has been dissolved in 50 ml water to produce a stock solution. What is the molarity of the stock solution? The assay mixture requires a final concentration of 200 µM isoleucine in 10 ml buffer. What volume of stock solution needs to be made up to 10 ml? (4 marks)

7. The optical density OD (or absorbance) at 600 nm of an overnight culture of the bacterium Vibrio fischeri is 0.4. What volume of this suspension do you need to make up to 100 ml with fresh medium to give an OD of 0.1 at 600 nm?

10. From a bottle of molten nutrient agar you need to pour 20 agar plates each containing 50 g ml-1 ampicillin. Each plate will contain ~ 25ml of nutrient agar, and you have a stock ampicillin antibiotic solution of 200 mg ml-1. What volume of ampicillin stock solution do you need to add to the bottle of molten agar before pouring the plates? (4 marks)

canalon
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Joined: Thu Feb 03, 2005 2:46 pm

### Re: Biochem Help?

Chemhalp wrote:What is the concentration in µg ml-1 of a solution of glycine of 23 g litre-1? (2 marks)
(Mol. wt of glycine = 75.07)

Thanks ?

Can I get some pointers on where to start

5. The following colony counts on nutrient agar plates were obtained from dilutions of an overnight culture of the bacterium Pseudomonas aeruginosa:

10-5 = >1000
10-6 = 622
10-7 = 74
10-8 = 14

Choose an appropriate colony count to determine how many colony-forming units (cfu) were present in 1 ml of the original culture (assuming that 100 l of the diluted culture was spread onto each agar plate)? (4 marks)

6. 50 mg of the amino acid isoleucine (molecular weight of 131.2) has been dissolved in 50 ml water to produce a stock solution. What is the molarity of the stock solution? The assay mixture requires a final concentration of 200 µM isoleucine in 10 ml buffer. What volume of stock solution needs to be made up to 10 ml? (4 marks)

7. The optical density OD (or absorbance) at 600 nm of an overnight culture of the bacterium Vibrio fischeri is 0.4. What volume of this suspension do you need to make up to 100 ml with fresh medium to give an OD of 0.1 at 600 nm?

10. From a bottle of molten nutrient agar you need to pour 20 agar plates each containing 50 g ml-1 ampicillin. Each plate will contain ~ 25ml of nutrient agar, and you have a stock ampicillin antibiotic solution of 200 mg ml-1. What volume of ampicillin stock solution do you need to add to the bottle of molten agar before pouring the plates? (4 marks)

Pointers:
1- convert g in mol with MW and you have your answer.
5- your prof should have told you about the best numbers for counting. Basically if a plate is too dense or too sparse the value cannot be trusted (in one case because there are too many to insure correct isolation, in the other because the effect of random distributio becomes too important too be negligible)
6- same as the first one
7- You have 4\$ one dollar bills, you owe me 1 dollar, how many bills do you give me? No seriously, this is as simple...
8- calculate the concentration in the bottle, the rest is just there to complicate things.
Patrick

Science has proof without any certainty. Creationists have certainty without
any proof. (Ashley Montague)

Chemhalp
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Pointers are awesome, exactly what I needed (still wanted to work them out myself)

For question 4 again...

conc = solute mass / ( mol wt * volume of solution)

so 23 / 75.01 = 0.3 gl-1

0.3/1000000 = 3x10^-6

?

Is that correct?

canalon
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Joined: Thu Feb 03, 2005 2:46 pm
Chemhalp wrote:Pointers are awesome, exactly what I needed (still wanted to work them out myself)

For question 4 again...

conc = solute mass / ( mol wt * volume of solution)

so 23 / 75.01 = 0.3 gl-1

0.3/1000000 = 3x10^-6

Is that correct?

23/75.07=0.31 mol.L-1 and then you convert to µmol/ml
Patrick

Science has proof without any certainty. Creationists have certainty without
any proof. (Ashley Montague)

Chemhalp
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Joined: Wed Nov 28, 2007 2:22 pm
Maybe I've just confused myself, but even after converting 0.31 mol.L-1 to µmol/ml how do I get to the µg/ml the question wanted?

Ah wait...

so mass/molar mass = 23/75.07 = 0.306mol
Concentration = mol/volume in L = 0.306mol/1 = 0.306mol/L
therefore 0.306mol/1000 (total ml in L) = 0.000306 (3.06 * 10^-4)
so 0.000306mol/mL
then weight = mol * molar mass = 0.000306*75.07 = 0.023g per mL

1000000μg = 1g

so 0.23g * 1000000 = 23,000μg/mL

Is that looking correct now?

blcr11
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Joined: Fri Mar 30, 2007 4:23 am
Why do you need to convert to molar units if your stock concentration is in mass/unit volume and you want the answer in a different set of mass/unit volume? Wouldn't you just want to convert directly from g/litre to ug/ml?

canalon
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Joined: Thu Feb 03, 2005 2:46 pm
blcr11 wrote:Why do you need to convert to molar units if your stock concentration is in mass/unit volume and you want the answer in a different set of mass/unit volume? Wouldn't you just want to convert directly from g/litre to ug/ml?

I think I might be responsible... I read µmol, not µg and give pointers fo this conversion. In this case going through the molarity is adding an useless layer of complexity. But the answer is correct
Patrick

Science has proof without any certainty. Creationists have certainty without
any proof. (Ashley Montague)

Chemhalp
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Joined: Wed Nov 28, 2007 2:22 pm
Haha, ty,

I've gone back through my notes and completed question 5 now, and from your pointers I've had a go at question 7.

If i'm right in the head i'm going to assume that I only need 25ml.

I have 100ml that absorbs 0.4 units
So I just divde that suspension into 4 parts (25ml) and dilute it with 75ml of fresh medium?

This would make it 4 times weaker (I assume) and mean that it only absorbs 0.1 units?

----------------

And then on that note, I also had a go at question 6 here how much I did...

I started off converting the 50mg to grams so 0.05g

So mass divided by molar mass = 0.05g/131.2 = 0.000381 mol

Now the molarity is mols per litre

So I have 0.000381 moll and 50ml

And this is where I lost myself last night (was too tired)

But I assumed that as 50ml is 0.05 litres, Would I just multiply both sides by 20 and then divide them?

So 0.000381 mol * 20 / 50ml * 20

Which is the same as just dividing them off the bat right? Yeah I got lost then...

Any help on where I went wrong there? Because I'm sure I got to the mol part right but couldn't work out last night how to get to the molarity.

7.62 × 10 ^ -6 was my answer anyway..

canalon
Inland Taipan
Posts: 3909
Joined: Thu Feb 03, 2005 2:46 pm
Question 7 is correct
Question 6: your calculation are correct until the last step. You have 3.81e-4 moles in 50 ml, so in 1L as you pointed out you would have 20 times more:
3.81e-4 x 20=7.62e-3 mol (per Liter)
so the molarity is 7.62e-3 M
No you just calculate the dilution as in question 7 to know how much of the stock solution you need to get a 200µM solution (200µM = 200 e-6 M = 2e-4 M)
Patrick

Science has proof without any certainty. Creationists have certainty without
any proof. (Ashley Montague)

Chemhalp
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Joined: Wed Nov 28, 2007 2:22 pm
Thankyou for the help with question 7 I had an idea it was a trick question and just looked harder but felt I was being stupid =/

As for question 6...

I have 50ml of stock solution with a molarity of 7.62e-3 (mol l-1 (i think))

And I need a final solution to have 2e-4M

Soo... I realise that the one I have is too big... But the question also asks me to make it so that I have a volume that is less than 10ml (From the question I have 50ml atm)

Now I think I've got the understanding for what the question wants... But I don't know where to go...

I worked out the molarity of my stock if I just took 10ml of it, and the result is:

7.62e-5

But this value is far far too low for the answer i'm supposed to be getting =/

I then worked out that I need 26.25 milliliters of my stock solution to have a molarity of 2e-4

What volume of stock solution needs to be made up to 10 ml

So my answer can't be 26.25 ml can it?

canalon
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Joined: Thu Feb 03, 2005 2:46 pm
No what you want is 10 ml of 200µmol per liter that is what µM means.

The traditional way of calculating is
Ci x Vi=Cf x Vf
Which basically says that you have the same number of molecule in the initial solution than in the final one (concentration x volume = #of molecules in moles)
Ci: initial concentration, in mol per liter, here 7.62e-3
Vi: volume of initial solution, that is what you want to calculate
Cf: final concentration, in mol per liter, here 200e-6
Vf: fianl volume, here 10ml
You just need to solve this for Vi and you have your answer
Patrick

Science has proof without any certainty. Creationists have certainty without
any proof. (Ashley Montague)

Chemhalp
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Joined: Wed Nov 28, 2007 2:22 pm
Been a little busy with other lecture work so have only just looked back at this.

Ci x Vi = Cf x Vf

so

Vi = Cf x Vf / Ci

2e-4 x 10 / 7.62e-3 =

0.264 ml?

Converting the 10ml into litres would just give me the answer in litres I thought, so I left it as ml =/

Think I did this right anyway

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